1 Determine \( f^{-1}(x) \) from first principles if \( f(x)=2 x^{2}-1 \) Determine: \( 7.2 .1 \quad \frac{d y}{d x} \) if \( y=[(2 x-1)(3 x+2)] \) \( 7.2 .2 \quad D_{x}\left[2 \sqrt{x}-\frac{2}{x^{8}}\right] \) .3 Determine the gradient of the graph of \( f(x)=-2 x^{3}-3 x+11 \) at \( x=-2 \).

To find the inverse \( f^{-1}(x) \) of the function \( f(x) = 2x^2 - 1 \), start by setting \( y = f(x) \), which gives us \( y = 2x^2 - 1 \). Next, rearranging the equation to solve for \( x \) gives us \( x^2 = \frac{y + 1}{2} \). Taking the square root results in \( x = \pm \sqrt{\frac{y + 1}{2}} \). Since \( f(x) \) is not one-to-one across all real numbers, we restrict \( x \) to non-negative values, thus \( f^{-1}(x) = \sqrt{\frac{x + 1}{2}} \) for \( x \geq -1 \). For the derivative \( \frac{dy}{dx} \) of the function \( y = (2x-1)(3x+2) \), we can use the product rule which states \( \frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \). Here, let \( u = (2x-1) \) and \( v = (3x+2) \). The derivative of \( u \) is \( 2 \) and that of \( v \) is \( 3 \). Thus, applying the product rule gives us: \[ \frac{dy}{dx} = (2x - 1)(3) + (3x + 2)(2) = 3(2x - 1) + 2(3x + 2). \] Now, calculate it: \[ = 6x - 3 + 6x + 4 = 12x + 1. \] Next, for \( D_x\left[2 \sqrt{x} - \frac{2}{x^8}\right] \), we differentiate each part separately. The derivative of \( 2\sqrt{x} \) is \( \frac{2}{2\sqrt{x}} = \frac{1}{\sqrt{x}} \) and the derivative of \( -\frac{2}{x^8} \) using the power rule is \( 16x^{-9} = \frac{16}{x^9} \). So we have: \[ D_x\left[2 \sqrt{x} - \frac{2}{x^8}\right] = \frac{1}{\sqrt{x}} + \frac{16}{x^9}. \] Lastly, to find the gradient of the graph \( f(x) = -2x^3 - 3x + 11 \) at \( x = -2 \), first differentiate the function: \[ f'(x) = -6x^2 - 3. \] Now, substitute \( x = -2 \): \[ f'(-2) = -6(-2)^2 - 3 = -6(4) - 3 = -24 - 3 = -27. \] Therefore, the gradient of the graph at \( x = -2 \) is \( -27 \).